This chapter is part of the verified current curriculum mapping for the academic session shown below.
Use this Class 10 Mathematics chapter hub to study Probability from one place. SaralStudy currently provides 2 visible NCERT solution entries on this page.
Review the SaralStudy solution entries currently available for this chapter.
Total no. of coins = 180
No. Of 50 paise coins = 100
No. Of 1-rupee coins = 50
No. Of 2-rupee coin = 20
No. Of 5-rupee coin = 10
(i) Probability of getting 50 paise coin = 100/180 = 5/9
(ii) No. Of not five-rupee coin = 170
Probability of not getting 5-rupee coin
= 170/180 = 17/18
No. of male fishes = 5
No. Of female fishes = 8
Total no. of fishes = 13
Probability of getting a male fish = 5/13
(i) Probability of getting number 8 = 1/8
(ii) Total odd numbers on the wheel = 4
Probability of getting an odd number = 4/8 = ½
(iii) Number greater than 2 = 6
Probability of getting no greater than 2 = 6/8 = ¾
(iv) Numbers less than 9 = 8
Probability of getting a no. Less than 9 = 8/8 = 1
Total no. Of possible outcomes = 6
(i) Prime numbers = 3
Probability of getting a prime no. = 3/6 = ½
(ii) Numbers between 2 and 6 = 3
Probability of getting a no. between 2 and 6
= 3/6 = ½
(iii) Odd numbers = 3
Probability of getting an odd no. = 3/6 = ½
Total no. Of cards = 52
(I) Numbers of king of red color = 2
Probability of getting king of red color =2/52=1/26
(ii) Number of face cards = 12
Probability of getting a face card = 12/52 = 3/13
(iii) Number of red face cards = 6
Probability of getting a face card = 6/52 = 3/26
(iv) Number of jack of hearts = 1
Probability of getting a jack of heart = 1/52
(v) Number of a spade = 13
Probability of getting a spade = 13/52 = ¼
(vi) Number of queens of diamond = 1
Probability of getting a queen of diamond = 1/52
Total no. Of cards = 5
(i) Number of cards of queen = 1
Probability of getting a queen = 1/5
(ii) Now, keeping the queen aside only four cards are left
So,
Total no. of outcome = 4
(a) Number of ace cards = 1
Probability of getting an ace = ¼
(b) Number of queen cards = 0
Probability of getting a card of queen = 0/4 = 0
Total no. Of pens = 144
Number of defective pens = 12
Number of good pens = 132
Probability of getting a good pen = 132/144 = 11/12
Total no. Of bulbs = 20
Number of defective bulbs = 4
Number of good bulbs = 16
(i) Probability of getting a defective bulb = 4/20 = 1/5
(ii) If one good bulb is kept aside,
Total no. Of bulbs = 19
Number of good bulb (not defective) = 15
Probability of getting not defective bulb = 15/19
Total number of discs = 90
(i) A 2-digit number discs = 81
Probability of getting two-digit number = 81/90 = 9/10
(ii) A perfect square number disc = 9
Probability of getting a perfect square number disc
= 9/90 = 1/9
(iii) A number divisible by 5 = 18
Probability of getting numbered divisible by 5
= 18/90 = 1/5
Total number of faces = 6
(i) A type of faces = 2
Probability of getting A type of face = 2/6 = ½
(ii) D type of face = 1
Probability of getting D type of faces = 1
Area of rectangle = 3×2 = 6 m2
Area of circle = π (½)2 = π/4 m2
Probability that the pie drops in the circle = (π/4) = = π/24
6
Total no. Of ball pens = 144
Number of defective ball pens = 20
Total no. Of good ball pens = 144 – 20 = 124
(i) Probability that she will get a good pen = 124/144 = 31/36
(ii) Probability that she will get a defective pen
= 20/144 = 5/36
(i) Total no. Of outcomes = 36
• (1, 2) and (2, 1) are events for getting a sum as 3
P (E) = 2/36 = 1/18
• (1, 3), (2, 2) and (3, 1) are the events of getting the Sum 4
P(E) = 3/36 = 1/12
• (1, 4), (2, 3), (3, 2) and (4, 1) are the events of getting the sum 5
P(E) = 4/36 = 1/9
• (1, 5), (2, 4), (3, 3), (4, 2) and (5, 1) are the events of Getting a sum 6
P(E) = 5/36
• (1, 6), (2, 5), (3, 4), (4, 3), (5, 2) and (6, 1) are the event of getting a sum 7
P(E) = 6/36 = 1/6
• (3, 6), (4, 5), (5, 4) and (6, 3) are the events of getting a sum 9
P(E) = 4/36 = 1/9
• (4, 6), (5, 5) and (6, 4) are the events of getting a sum 10
P(E) = 3/36 = 1/12
• (5, 6), (6, 5) are the events of getting a sum 11
P(E) = 2/36 =1/18
(ii) No, the eleven sum is not equally likely.
Total no. Of outcomes = 8
Number of outcomes if Hanif wins = 2
P(E) = 2/8 = ¼
P(not E) = 1 – ¼ = ¾
Total no. Of possible outcomes = 36
(i) 5 will not come either up either time = 25
P(E) = 25/36
(ii) 5 will come up at least time = 11
P(E) = 11/36
Total no. Of outcomes = 6
(i) P (two tails) = ¼
P (two heads) = ¼
P(one head and one tail) = 2/4 =½
So, this argument is incorrect.
(ii) P (odd no.) = 3/6 = ½
P (even no.) = 3/6 = ½
So, this statement is correct.
Because the outcomes of a coin head or tail are equally likely. So, this is the fair way to decide which team get the ball at the beginning.
- 1.5 because probability of an event always lies between 0 and 1.
P (E) + P (not E) = 1
0.05 + P (not E) = 1
P (Not E) = 1 – 0.05 = 0.95
P (E) + P (not E) =1
P (E) + 0.992 = 1
P (E) = 1 - 0.992 = 0.008
(i) Total no. balls = 8
Red balls = 3
Probability of red balls = 3/8
(ii) Not red balls = 5
Probability of not getting red balls = 5/8
Total no. of marbles = 17
(i) Total no. of red marbles = 5
Probability of getting red marbles = 5/17
(ii) Total no. of white marbles = 8
Probability of getting white marbles = 8/17
(iii) No. of not green marbles
= total no. of marbles – no. of green marbles
= 17 – 4= 13
Probability of getting not green marbles = 13/17
Since the number of days on which they both visit is 5 (Tuesday, Wednesday, Thurday, Friday, Saturday) they both can visit in 5 ways.
Therefore, total no of possible outcomes 5×5 = 25
(i) When the both visits the same day
Total no of favourable outcomes = 5 (Tuesday, Tuesday)
(Wednesday, Wednesday) (Thursday,Thursday) (Friday, Friday) (Saturday, Saturday)
So,P(both visiting the same day) = 5/25 = 1/5
(ii) When they visits on consecutive days 8 (Tuesday, Wednesday)
(Wednesday, Thursday) (Thursday, Friday) (Friday, Saturday) (Saturday, Friday) (Friday, Thursday)
(Thursday, Wednesday) and (Wednesday, Tuesday)
P (both visitng the consecutive days) = 8/25
(iii) when they visits the different days
P(E) = 1 – P (visits the Same day)
= 1 – 1/5 =4/5
|
+ |
1 |
2 |
3 |
4 |
5 |
6 |
|
1 |
2 |
3 |
3 |
4 |
4 |
7 |
|
2 |
3 |
4 |
4 |
5 |
5 |
8 |
|
3 |
3 |
4 |
4 |
5 |
5 |
8 |
|
4 |
4 |
5 |
5 |
6 |
6 |
9 |
|
5 |
4 |
5 |
5 |
6 |
6 |
9 |
|
6 |
7 |
8 |
8 |
9 |
9 |
12 |
Total no of outcomes = 6×6 = 36
(i) Even
Total no of favourable outcomes = 18
P(Even) = 18/36 = ½
(ii) Sum is 6
Total no of favourable outcomes = 4
P (sum is 6) = 4/36 =1/9
(iii) Sum is at least 6
Total no of favourable outcomes = 15
P(sum is at least 6) = 15/36 = 5/12
Let the number of blue balls = x
Number of red balls = 5
Therefore, total no. Of balls = 5 + x
P(E) of drawing a blue ball = [x / (5 + x)] ...........(I)
P(E) of drawing a red ball = [5 / (5 + x)]
Acc. to question,
P(E)B = 2 P(E)R
[x / (5 + x)] = 2 [5 / (5 + x)]
x = 10
Total no of balls = 12
Let the no. of black balls = x
P(E) of getting a black ball = [ x / 12]
(ii) When 6 more black balls added to bag
Total no of balls = 12 + 6 = 18
No of black balls = x + 6
P (black ball) = [(x + 6) / 18]
Acc. to question,
Before After
2 P(E)B = P(E)B
2 [x / 12] = [(x + 6) / 18]
x/6 = x + 6/18
x + 6 = 3x
2x = 6 => x =3
Total no of marbles in the jar (green + blue) = 24
Let the no of green marbles be = x
Therefore, no of blue marbles left = 24 – x
P(E) of marble to be green = 2/3 {Given} …....(I)
Acc. to question
P(E) of green marble = x/24 …..........(ii)
Equating equation (I) and (ii)
2/3 = x/24
x = 16
No of green marbles = 16
Hence, no of blue marbles = 24- 16 = 8
Use the available NCERT textbook PDF alongside the chapter solutions.
Use the mapped NCERT Exemplar resource for additional chapter practice.
Current ExemplarCurrent and historical curriculum records are shown separately so removed material is not mixed into current practice.
Fresh educational articles and updates.
Most-read articles from SaralStudy.
Fast access to frequently used study resources.
Recently viewed articles based on SaralStudy activity.
(B) –1.5 (C) 15% (D) 0.7